amendola l. dark energy

• Gravity, expansion and• Dark energy: linear and non-linear properties... • Gravity is always e.• Aristotle's problem: how to avoid that thesky falls on our head?... • Gravity is always

Amendola, INAF/OARamendolamporzio.astro.it

7th er 2005

• Gravity, expansion and

• Dark energy: linear and non-linear properties

• Gravity is always e.

• Aristotle's problem: how to avoid that thesky falls on our head?

• Gravity is always e.

• Einstein's answer: to avoid (to makethe universe stable) it is to intro-

a form of repulsive gravity, by modifyingthe equations of General Relativity

R µν − 1

2 Rg µν = 8πGT µν gravity matter

pends on ρ, p, u µ = dx µ /ds, g µν and to thislimit in the Minkowski

T µν = (ρ + p)u µ u ν − pg µν (2)

Einstein's equations are only when a

re-lation between p and ρ is given: the equation ofstate

R µν − 1

2 Rg µν = 8πGT µν

we add whatever rank-2 tensor

vari-antly ed, e.g any E µ ν h that

• instead the fourth we

add a (small) term Λg µν and rewrite theequations as

R µν − 1 2 Rg µν − Λg µν = 8πGT µν

• The big idea of the latter years has been to

move the new term from right to left

p Λ = − 8π Λ , ρ Λ = Λ

8π

nega-tive pressure (if Λ > 0 )

• Intro the equation of state

p = ρ/3 → w = 1/3

of isotropy, the spatial ds 2 3 = σ ij dx i dx j

depend only on |r| and on dx 2 + dy 2 + dz 2 =

dr 2 + r 2 (dθ 2 + sin 2 θdφ) Then

ds 2 3 = a 2 (t)λ 2 (r)[dr 2 + r 2 (dθ 2 + sin 2 θdφ)]

or, redening r,

ds 2 3 = a 2 (t)[λ ′2 (r ′ )dr ′2 + r ′2 (dθ 2 + sin 2 θdφ)] (3)

so we are left with a single unknown λ(r)

But we have still to impose homogeneity How?

Embedding a 3D homogeneous hypersphere in a

1 − kr 2

So anlly we obtain the most general homogenous

and isotr , the of F

riedmann-Robertson-ds 2 = dt 2 −a 2 (t)[ dr

2

1 − kr 2 +r 2 (dθ 2 +sin 2 θdφ 2 )] (6)

• What has to do a negative pressure with

and the energy density (here we put for

y k = 0 and always assume a 0 = 1)

w < −1/3

then we get expansion

with w < −1/3) the expansion

→ repulsive gravity We this

hypo-uid Dark Energy

• Consider now only the t

• Generally speaking, there are at least three

onents (plus ature) so that

Ordinary matter (baryons plus dark matter)

es energy during expansion, so that

we have three dierent behaviors

ρ γ ∼ a −4

ρ M ∼ a −3

ρ k ≡ a k 2 ∼ a −2

ρ Λ ∼ a 0

• In general, therefore, we have

rad → matter → curvature → cosm.const.

• Think of a eld, eg a eld, as a series

always put to zero

• Quan , however, the state of mum is not at zero energy but rather

mini-E 0 = 1

2 ~ ω

• Therefore, for a eld, the total zero-point

of size L, we obtain dn i = dk i L/2π modes inthe range dk i, so that

k max is then

ρ vacuum = lim E

L 3 = ~ k

4 max

16π 2

• The energy diverges at the high

(ultraviolet div We must supposethen that there is k max beyond h a new

in modies the system

• The problem is, h k max ? If we assume

as limit the k energy

orders of magnitude!

• This fundamental problem is stillopen

• We have seen that the Friedmann equationhas the form

H 2 = 8π

3 (ρ M + ρ Λ + ρ k + ?)

• Suppose we know nothing of the matter

tent of the universe How to study the

ture of the ?

that a smooth onent, as the Λ, has

no lo observational The Poisson tion remains invaried, b in linear GR thePoisson equation

• What we really observe in is lightfrom and from kgrounds.

• How do we these observables to

quantities like ρ m , ρ γ , k, a(t), H 0

• First, dene

Ω M = 8πρ 0

3H 0 2 , Ω Λ =

8πρ Λ 3H 0 2 , Ω k =

8πk 3H 0 2

and note that

1 = Ω M + Ω Λ + Ω k

so rewrite Friedman equation as (a 0 = 1)

H 2 = H 2 (Ω a −3 + Ω a 0 + Ω a −2 )

• Then, generalize it to several onents:

• Unknown quantities: H 0 , Ω i , w i to be mined using:

deter-1 Angular positions of e.g galaxies:

θ i , ϕ i

2 Redshifts: z i

3 Apparent magnitudes: m i

4 Ages of stars

5 kground radiation e.g CMB: ∆T /T

• BASIC RELATION

a = 1+z 1

• The age of the universe be fromthe Friedmann equation:

 da dt

 2

= H 0 2 a 2 E(a) 2

we get

 dz

Ω tot = Ω M + Ω Λ is xed, is to the

age

• From at Friedmann's

ds 2 = c 2 dt 2 − a 2 dr 2

and integrating along the null geo weget the proper h is what youwould measure with xed rods

r =

Z cdt a(t) = c

Z da

˙aa = c

Z

dz H(z)

→ generalized Hubble law: measuring

dis-means measuring

• If we the energy L emitted by a

at proper r with ux f

arriv-d(z) h that

4πr 2 (1 + z) 2 = L

4πd 2

• The two extra of 1 + z take into

t the loss of energy due to redshift andthe spread of energy due to the relative di-latation of the emission time versus observer'stime We get

d(z) = r(1 + z) = cH 0 −1 (1 + z)

Z z 1

0

dz E(z)

where cH 0 −1 = 100hkm/sec/M pc 300.000km/sec = 3000h −1 M pc

• Remember our

E 2 (z) = Ω M (1 + z) 3 + Ω Λ + Ω K (1 + z) 2

• The luminosity therefore depends upon

lu-minosity of the standard

• Suppose we have a of known absoluteluminosity M = −2.5 log L + const Thenone denes instead of the ux f an apparentmagnitude m = −2.5 log f + const as

W L

z=4

d(z; Ω , Ω )

• There exist standard in nature ?

• The best h thing so far are supernovaeIa

• This hypothesis be tested and

through a lo sample whose weknow by other means

• Then, we m obs (z) with